If `A=[{:(alpha,beta),(gamma,alpha):}]` is such that `A^2=I` , then `1+alpha^2+betagamma=0` (b) `1-alpha^2+betagamma=0` (c) `1-alpha^2-betagamma=0` (d) `1+alpha^2-betagamma=0`

Options

A. `1+alpha^(2)+betagamma =0`
B. `1-alpha^(2) +beta gamma =0`
C. `1-alpha^(2)-beta gamma=0`
D. `1+alpha^(2)-beta gamma =0`

Explanation

To solve the problem, we need to find the condition on the parameters \( \alpha \), \( \beta \), and \( \gamma \) such that the matrix \( A \) defined as

\[ A = \begin{pmatrix} \alpha & \beta \\ \gamma & \alpha \end{pmatrix} \]

satisfies the equation \( A^2 = I \), where \( I \) is the identity matrix.

### Step-by-Step Solution:

**Step 1: Compute \( A^2 \)**

To find \( A^2 \), we multiply the matrix \( A \) by itself:

\[ A^2 = A \cdot A = \begin{pmatrix} \alpha & \beta \\ \gamma & \alpha \end{pmatrix} \cdot \begin{pmatrix} \alpha & \beta \\ \gamma & \alpha \end{pmatrix} \]

Calculating this gives:

\[ A^2 = \begin{pmatrix} \alpha^2 + \beta \gamma & \alpha \beta + \beta \alpha \\ \gamma \alpha + \alpha \gamma & \gamma \beta + \alpha^2 \end{pmatrix} \]

This simplifies to:

\[ A^2 = \begin{pmatrix} \alpha^2 + \beta \gamma & 2\alpha \beta \\ 2\gamma \alpha & \gamma \beta + \alpha^2 \end{pmatrix} \]

**Step 2: Set \( A^2 \) equal to the identity matrix \( I \)**

The identity matrix \( I \) is given by:

\[ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

Setting \( A^2 \) equal to \( I \):

\[ \begin{pmatrix} \alpha^2 + \beta \gamma & 2\alpha \beta \\ 2\gamma \alpha & \gamma \beta + \alpha^2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

This gives us the following equations:

1. \( \alpha^2 + \beta \gamma = 1 \)
2. \( 2\alpha \beta = 0 \)
3. \( 2\gamma \alpha = 0 \)
4. \( \gamma \beta + \alpha^2 = 1 \)

**Step 3: Analyze the equations**

From the second equation \( 2\alpha \beta = 0 \), we can conclude that either \( \alpha = 0 \) or \( \beta = 0 \).

From the third equation \( 2\gamma \alpha = 0 \), we conclude that either \( \gamma = 0 \) or \( \alpha = 0 \).

**Case 1: If \( \alpha = 0 \)**

Substituting \( \alpha = 0 \) into the first equation gives:

\[ 0 + \beta \gamma = 1 \implies \beta \gamma = 1 \]

Substituting \( \alpha = 0 \) into the fourth equation gives:

\[ \gamma \beta + 0 = 1 \implies \beta \gamma = 1 \]

Both equations are consistent.

**Case 2: If \( \beta = 0 \)**

Substituting \( \beta = 0 \) into the first equation gives:

\[ \alpha^2 + 0 = 1 \implies \alpha^2 = 1 \implies \alpha = \pm 1 \]

Substituting \( \beta = 0 \) into the fourth equation gives:

\[ \gamma \cdot 0 + \alpha^2 = 1 \implies \alpha^2 = 1 \]

Both equations are consistent.

**Step 4: Conclusion**

From the equations derived, we can summarize:

1. If \( \alpha = 0 \), then \( \beta \gamma = 1 \).
2. If \( \beta = 0 \), then \( \alpha^2 = 1 \).

Thus, we can conclude that:

\[ 1 - \alpha^2 - \beta \gamma = 0 \]

This matches option (c): \( 1 - \alpha^2 - \beta \gamma = 0 \).

### Final Answer:
(c) \( 1 - \alpha^2 - \beta \gamma = 0 \)

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